[tex](a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n[/tex](adsbygoogle = window.adsbygoogle || []).push({});

all above is just done by using that the order of the factors that you multiply does not matter

we have proven that

[tex](a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n[/tex]

for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:

[tex]\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}} [/tex]

to become right side by only using roths and powers of integers.

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# Whole roots and powers

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